To solve the integral:

$I={\int }_0^{\pi }\log \left(\sin \right(x)dx$

We need to proceed carefully. This integral is famous for having a nontrivial solution, and the derivation involves several steps, including symmetry considerations and the use of advanced techniques. Below is a detailed step-by-step solution.

Step 1: Symmetry of the Function

The sine function is symmetric about $\mathrm{\pi }/2$ in the interval $0$ to $\pi$. This symmetry allows us to rewrite the integral over half the interval:

$I=2{\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx$

This step reduces the limits of integration from 0 to π/2 instead of 0 to π, making the next steps more manageable.

Step 2: Use of a Known Identity

There is a standard trick used to deal with integrals involving logarithms and trigonometric functions. We introduce a known identity for integrals involving logarithmic functions:

${\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx={\int }_0^{\frac{\pi }{2}}\log (\cos \left(x\right)dx$

This symmetry between sine and cosine integrals will be useful later.

Step 3: Combining the Integrals of log(sin(x)) and log(cos(x))

Now, consider the sum of these two integrals:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\sin \right(x)dx+{\int }_0^{\frac{\pi }{2}}\log (\cos \left(x\right)dx$

Using the identity $\mathrm{sin(x)\; cos(x)\; =\; (1/2)sin(2x)}$, we get:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\mathrm{sin(2x)}\right)dx$

This can be expanded as:

$S={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\right)dx+{\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{sin(2x)}\right)dx$

Step 4: Simplifying the Result

The first term is straightforward:

${\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{2}\right)dx=-\frac{\pi }{2}\mathrm{log(2)}$

The second term involves a change of variables. Let $u = my aniket roy August 28, 2023 Tags: aniket roy Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam nonummy nibh euismod tincidunt ut laoreet dolore magna Veniam, quis nostrud exerci tation ullamcorper suscipit lobortis nisl ut aliquip ex ea commodo consequat. Contact Us Email : aniketkroy@gmail.com Send Message Search Powered by Blogger. Blog Archive August 2023 (2) Report Abuse $